1. A shifted uniform series starts at a time other than year 1 When using P/A or A/P, P is always one year ahead of first A When using F/A or A/F, F is in same year as last A Solution: (1) Use P/A factor with n=5 to get P in year 1: (2) Use P/F factor with n=1 to move P back to year 0: P 1 = 10,000(P/A,10%,5) =10,000(3.7908) = $37,908 P 0 = P 1 (P/F,10%1) = 37,908(0.9091) = $34,462 Answer is (c) P=? A=10,000 0 1 2 3 4 5 6 i=10% Example: The present worth of the cash flow shown below at i=10% is: (a) $25,304 (b) $29,562 (c)$34,462 (d) $37,908

2. For A/P factor, find P in yr 0, then annualize with (A/P,12%,7) For A/F factor, find F in yr 7, then annualize with (A/F,12%7) Using A/F: A=[10,000(F/A,12%,6) + 6,000(F/P,12%,2)](A/F,12%,7) Solution: The annual worth in yrs 1-7 can be found using either the A/P or A/F factors =[10,000(8.1152) + 6,000(1.2544)](0.09912) =$8,789.80 Answer is (a) These cash flows require the use of multiple factors A=10,000 i= 12% 0 1 2 3 4 5 6 7 6000 Example: The equivalent annual worth in yrs 1-7 for the cash flow shown below at i=12% per year is: (a) $8790 (b) $9530 (c) $10,330 (d) $11,780

3. Shifted gradients begin at a time other than between periods 1& 2Must use multiple factors to find P in year 0 Arithmetic

4. Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397

5. Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 0 1 2 3 10 45 60 65 70 95 The cash flow diagram is as follows: Cash Flow Diagram i=12%

6. First find P 2 for the gradient ($5) and its base amount ($60) in year 2: P 2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 0 1 2 3 10 45 P=? 60 65 70 95 01238 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 The cash flow diagram is as follows: i=12%

7. First find P 2 for the base amount ($60) & gradient ($5) in year 2: P 2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 Next, move P 2 back to year 0: P 0 = P 2 (P/F,12%,2) = $295.29 0 1 2 3 10 45 P=? 60 65 70 95 01238 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 The cash flow diagram is as follows: i=12%

8. First find P 2 for the base amount ($60) & gradient ($5) in year 2: P 2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 Next, move P 2 back to year 0: P 0 = P 2 (P/F,12%,2) = $295.29 Next, find P A for the $60 amounts of years 1 & 2: P A = 60(P/A,12%,2) = $101.41 0 1 2 3 10 45 P=? 60 65 70 95 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 The cash flow diagram is as follows: i=12%

9. First find P 2 for the base amount ($60) & gradient ($5) in year 2: P 2 = 60(P/A,12%,8) + 5(P/G,12%,8) = $370.41 Next, move P 2 back to year 0: P 0 = P 2 (P/F,12%,2) = $295.29 Next, find P A for the $60 amounts of years 1 & 2: P A = 60(P/A,12%,2) = $101.41 Finally, add P 0 & P A to get P T in year 0: A P T = P 0 + P A = $396.70 Answer is (d) 0 1 2 3 10 45 P T = ? 60 65 70 95 Example : John Deere expects the cost of a certain tractor part to increase by $5 /yr beginning 4 years from now. If the cost in years 1-3 is $60, the present worth of the cost thru year 10 at an interest rate of 12%/yr is closest to: (a)$101 (b)$295 (c)$370 (d)$397 The cash flow diagram is as follows: i=12%

10. Shifted gradients begin at a time other than between periods 1& 2 Geometric Equation yields P for all cash flow(i.e. base amt is included)For negative gradient, change signs in front of both g’s P=A{1-[(1+g)/(1+i)] n /(i-g)}

11. For arithmetic gradients, change sign in front of G term from + to - For goemetric gradients, change signs in front of both g’s All other procedures are the same as for positive gradients